当我的变量被定义时，我总是收到一个 undefined variable 错误。我尝试了 Stack Overflow 的多种不同建议，但没有一个对我有用。此页面的目的是编辑 PHP 表格中的内容。
我曾尝试以本网站上建议的不同方式重新定义我的变量。例如，我尝试为“myID”添加“mysql_fetch_object”。我也尝试过使用一个新变量来定义“myID”($newID = mysql_fetch_object($myID))。
<?php
// Server credentials
$servername = "***";
$username = "***";
$password = "***";
$dbname = "***";
// Creating mysql connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Checking mysql connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// Writing a mysql query to retrieve data
// ERROR HERE
$myID = $_GET['Event_id'];
$sql = "SELECT Event_id, title, event_date, location, description, status FROM events WHERE Event_id = $myID";
$result = $conn->query($sql);
//if ($result->num_rows == 1) {
// ERROR HERE
//if ($result) {
// Show each data returned by mysql
while($row = $result->fetch_assoc()) {
if(isset($_POST['submitc'])) {
changeform();
} else{
header('location: events.php');
}
function changeform(){
$title= mysqli_real_escape_string($conn, $_POST['title']);
$event_date = mysqli_real_escape_string($conn, $_POST['event_date']);
$location = mysqli_real_escape_string($conn, $_POST['location']);
$description = mysqli_real_escape_string($conn, $_POST['description']);
if(!empty($title)) {
mysqli_query($conn, "UPDATE events SET title = '$title' WHERE Event_id = '$myID'");
} else{
}
if(!empty($event_date)) {
mysqli_query($conn, "UPDATE events SET event_date = '$event_date' WHERE Event_id = '$myID'");
} else{
}
if(!empty($location)) {
mysqli_query($conn, "UPDATE events SET location = '$location' WHERE Event_id = '$myID'");
} else{
}
if(!empty($description)) {
mysqli_query($conn, "UPDATE events SET description = '$description' WHERE Event_id = '$myID'");
} else{
}
}
?>
<html lang="en">
<head>
<meta charset="utf-8">
<title>I491 MakeUp</title>
<link rel="stylesheet" href="eventcss.css">
</head>
<header>
<div id="text">
<!-- script for header animation-->
<script class="logo" type="text/javascript">
var i=0, text;
text = "I491 Make-Up"
function typing(){
if(i<text.length){
document.getElementById("text").innerHTML += text.charAt(i);
i++;
setTimeout(typing,50);
}
}
typing();
</script>
</div>
</header>
<body>
<div class="edit_part">
<form method="post" action="editevent.php">
<p> Title: <?php echo $row["title"]; ?>
<input type="text" placeholder="Title" name="title" value="<?php echo $title; ?>">
</p>
<p> Date: <?php echo $row["event_date"]; ?>
<input type="text" placeholder="Date" name="date" value="<?php echo $event_date; ?>">
</p>
<p> Location: <?php echo $row["location"]; ?>
<input type="text" placeholder="Location" name="location" value="<?php echo $location; ?>">
</p>
<p>
Description: <?php echo $row["description"]; ?>
<input type="text" placeholder="Description" name="description" value="<?php echo $description; ?>">
</p>
<button type="submit" name="submitc" value="1-or-anything">
Submit Changes
</button>
</form>
</div>
</body>
<?php
}
//}
// else {
// echo "0 results";
// }
// Closing mysql connection
$conn->close();
我的期望是能够将 PHP 表更改为用户通过单击提交按钮输入的输入。但如果用户未输入任何内容，也不会更改输入。
最佳答案
您正在定义 function():
function changeform() {
// Your func body, where yo're using $myID now
}
您的函数无法在函数体外部看到变量 $myID。函数可以看到变量，如果变量定义在函数体中:
funciton changeForm() {
$myId = 1;
}
或者您正在从另一个作用域传输到函数参数:
$myId = 1;
function changeForm($myId) {
$myId++;
}
changeForm($myId);
我建议您阅读有关 variable scope 的文章
关于php - 我收到一个 undefined variable ，但我的变量是在从另一个页面接收主变量 (myID) 时定义的，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/57815664/